The tangent line to the graph of function $f$ at the point $(5,-7)$ passes through the point $(1,-1)$. Find $f'(5)$. $f'(5)=$
Solution: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $f'(5)$ gives the slope of the tangent line to the graph of $f$ where $x=5$, which is the point $(5,-7)$. We know this line passes through $(5,-7)$, and we are also given that it passes through $(1,-1)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{-1-(-7)}{1-5} \\\\ &=\dfrac{6}{-4} \\\\ &=-\dfrac32 \end{aligned}$ In conclusion, $f'(5)=-\dfrac{3}{2}$.